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Two cells of same emf epsilon but of different internalresistances r_1 and r_2 are connected in series with an external resistance R. The potential drop across the first cell is found to be zero. The external resistance R is |
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Answer» `r_1+r_2` `I=(EPSILON+epsilon)/(r_1+r_2+R)=(2epsilon)/(r_1+r_2+R)` As TERMINAL potential drop across `1^(st)` CELL is zero , hence `V_1=epsilon-Ir_1=epsilon-(2epsilon)/((r_1+r_2+r))r_1=0` `RARR epsilon=(2epsilonr_1)/((r_1+r_2+R))` or `r_1+r_2+R=2r_1` or `R=(r_1-r_2)` 
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